6) Two ice skaters suddenly push off against one another starting from a stationary position. The 45-kg skater acquires a speed of 0.375 m/s relative to the ice. What speed does the 60-kg skater acquire relative to the ice?
.281m/s
To find the speed of the 60-kg skater relative to the ice, we can use the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum before the ice skaters pushed off against each other is equal to the total momentum after they pushed off.
The momentum of an object can be calculated by multiplying its mass by its velocity. Let's denote the mass of the 45-kg skater as m1 and the mass of the 60-kg skater as m2. The initial velocity of both skaters is zero, as they start from a stationary position. The final velocity of the 45-kg skater is 0.375 m/s relative to the ice, which we can denote as v1. We need to find the final velocity of the 60-kg skater, which we can denote as v2.
Using the principle of conservation of momentum, we can write the following equation:
(m1 * 0) + (m2 * 0) = (m1 * v1) + (m2 * v2)
Since the initial velocities are both zero, they do not contribute to the total momentum. Rearranging the equation gives:
m2 * v2 = -m1 * v1
Now, we can substitute the given values into the equation:
60 kg * v2 = -(45 kg * 0.375 m/s)
Simplifying further:
60 kg * v2 = -16.875 kg⋅m/s
Finally, we can solve for v2:
v2 = (-16.875 kg⋅m/s) / (60 kg)
v2 ≈ -0.28125 m/s
The speed of the 60-kg skater relative to the ice is approximately 0.28125 m/s in the opposite direction to the 45-kg skater's velocity.